# Commutators with a density matrix

The equation describing the evolution of our system is as follows:

$\dot{\rho} = u_1(t)(a^\dagger a \rho - 2a\rho a^\dagger +\rho a^\dagger a) + u_2(t)(a a^\dagger \rho - 2a^\dagger\rho a +\rho a a^\dagger)$

where $\rho$ is the density matrix and $a$ is the photon annihilation operator.

I need to rearrange this equation to look something akin to:

$\dot{\rho} = u_1(t) \hat{A} \rho + u_2(t) \hat{B}\rho$

How do I handle the commutators and the like in order to rearrange it?

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If the symbols $\hat A,\hat B$ mean some functionals of $a,a^\dagger$ which don't depend on $\rho$, your problem has no solution. You simply cannot rewrite objects such as $\rho H$ in the form $H' \rho$; matrices and operators don't commute with each other and their refusal to commute isn't a formality – it's a fundamental fact.

A proof that it can't be done?

The right hand side of your desired equation has the form $X\rho$ where $X=u_1(t)A+u_2(t)B$. Imagine that at $t=0$, there exists a state $|\psi\rangle$ that is annihilated by $\rho$ i.e. $$\rho |\psi\rangle = 0$$ Then your desired form of the equation implies $$\dot \rho |\psi\rangle = X\rho|\psi\rangle = 0$$ as well, regardless of the choice of $X$. However, it's clear that the right – first – equation you wrote at the top allows $\dot\rho |\psi\rangle$ to be nonzero. In fact, it will generically be nonzero because states such as $a|\psi\rangle$ and $a^\dagger \psi\rangle$ that appear as the "right side factors" in various terms are generically nonzero and the extra $\rho$ in front of them no longer annihilates them.

I am just saying that if a particular state $|\psi\rangle$ has a vanishing probability to be realized at $t=0$ according to the density matrix, it doesn't mean that it will have a vanishing probability at all later times. However, your desired form of the equation would imply that it stays zero forever. It can't be right so your equation cannot be equivalent to the original one regardless of the choice of $X$.

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The equation of motion looks like one from quantum optics, or spin. That means your $\rho$ is a 2x2 matrix, and you can decompose it as a sum of Pauli matrices. In this case, you can concretely work out the commutator with the raising and lower operators $a$ and $a^\dagger$. –  genneth May 30 '12 at 9:31
For the $2\times 2$ or any other case, you can calculate the commutators but for different terms in $\rho$, the commutators will have different forms, so you can't rewrite all of them in the universal form $X\rho$: the Heisenberg equation isn't "right side linear" in $\rho$. Moreover, it's totally obvious that the OP's problem wasn't about $2\times 2$ matrices; it included creation and annihilation operators so it was at least the quantum harmonic oscillator which is already infinite-dimensional. –  Luboš Motl May 30 '12 at 9:59
Agreed about the non-linearity in $\rho$. But it is often the case that one neglects high occupancy in quantum optics problems, which renders the Hilbert space finite again. And in extreme cases, 2. –  genneth May 30 '12 at 10:30
Excellent - if you supplement your suggestion with an explicit answer. –  Luboš Motl May 30 '12 at 10:31
Perhaps you misread: I agree that you can't write it in the form the OP wanted! I just wanted to say that perhaps he can still work out the commutator. –  genneth May 30 '12 at 10:33